If you want to obtain the maximum power transfer between a generator and a
load, and if all the pieces behave like ideal linear components, then you
need to match the impedance of the two. But this is not what you are doing
here. First, you want to obtain the maximum power out without burning up
you output device, and second, a class-C output stage isn't a linear
circuit!

What this means is that in general your transmitter, like your wall socket,
can deliver more than it's rated power if you put the right load on it.
Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it. Your
circuit doesn't act exactly like that, even if it doesn't follow the same
curve for a generator with a linear impedance. I suspect that with your
circuit the 7-pole filter is evening things out quite a bit. There's also a
good chance that the "strange results" you see with a load below 10 ohms are
the final amplifier going unstable and oscillating, or your output
transistor heating up and changing characteristics on you.

So what it boils down to is that it is very important that your output stage
_sees_ the right impedance, but you shouldn't expect the output stage to
_deliver_ the same impedance that it needs to see. As long as you're
getting power out and your output transistor isn't letting all the smoke out
then you're fine.

"Gary Morton" wrote in message
...
Over the Xmas break I constructed a simple 2 transistor QRP transmitter
using
the Tuna Tin II design.

Using a power meter and a 50 ohm dummy load it appears to show around
200mW, a
little less than the article suggested. Then again I had substituted the
final
stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too
sure
what exact difference this will make.

In the article it says that the 200 ohm output impedance of the final
stage is
transformed down to around 50 ohm (using a 2:1 turns ratio untuned
transformer wound on a toroid).

I have added a 7th order low pass filter to reduce the harmonics.

I am interested in verifying the output impedance.

If I understand the theory, when the output is terminated with a resistor
which matches the output impedance then the power transferred to the load
will
be maximised.

I set about loading the output will a series of resistors and measuring
the
peak to peak voltage across them in order to calculate the r.m.s. voltage
and
hence the power dissapation.

Resistors of value less than 10 ohm gave strange results.

load r.m.s. power (V*V/R)
voltage

10 1.272 0.162
15 1.767 0.208
27 3.005 0.334
33 3.253 0.321
39 3.676 0.346
47 4.066 0.352
56 4.384 0.343
68 4.666 0.320
100 5.303 0.281

The numbers work reasonably well and point towards 50 ohms-ish.

The power value looks too high - so I might have made a mis-calculation
somewhere!

Can anyone pass comments on the above?

I'm interested in where the 200 ohm figure comes from.

Any suggestions as to other methods of measuring the output impedance?

regards...

--Gary (M1GRY)

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:

Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.

??? Shirley, some mistake.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

Paul Burridge wrote:

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:



Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.


Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.

No, that isn't the case. From AN721:

"The load value is primarily dictated by the required output power and
the peak voltage; it is not matched to the output impedance of
the device."

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can
fool yourself into thinking it is, but it's not. (2) the typical RF output
stage, sans output filter, which consists of a transistor fed by a choke or
transformer and which pulls the collector voltage nearly down to the emitter
acts like an RF voltage source with a fairly low series resistance. I've
_done_ this. I've _measured_ this. I've _burnt_ transistors.

You _can_ get much more power from your output stage than it's "normal"
output power, for that brief (and often shining) moment before the output
transistor lets the smoke out and reverts to the sand and basic
petrochemicals from which it was originally made. This is how you know that
you aren't running it at the Rs = Rl point -- rather, you're running it at
the point where the junction temperature of the transistor is kept to a
level that will let it stay a transistor for a satisfying amount of time.

This is why in my original reply I compared the output amp to a wall socket.
The equivalent Rs of your average wall socket is probably on the order 1/2
to 4 ohms (on the low end for 120V, high for 230). Since I live in the US
and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs
with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit
is only is only good for 15 amps and I'm asking for 120?!? Dang! What
happened to the lights?

So you see, every time you plug anything into the wall, you're failing to
match Rs with Rl, and that's a _very_ good thing. It's for exactly the same
reason that you don't match Rs with Rl in your output stage.

"Paul Burridge" wrote in message
...
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:
-- snip --
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
-
Winston Churchill

Paul Burridge wrote:

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:



Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.


Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.

No, that isn't the case. From AN721:

"The load value is primarily dictated by the required output power and
the peak voltage; it is not matched to the output impedance of
the device."

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can
fool yourself into thinking it is, but it's not. (2) the typical RF output
stage, sans output filter, which consists of a transistor fed by a choke or
transformer and which pulls the collector voltage nearly down to the emitter
acts like an RF voltage source with a fairly low series resistance. I've
_done_ this. I've _measured_ this. I've _burnt_ transistors.

You _can_ get much more power from your output stage than it's "normal"
output power, for that brief (and often shining) moment before the output
transistor lets the smoke out and reverts to the sand and basic
petrochemicals from which it was originally made. This is how you know that
you aren't running it at the Rs = Rl point -- rather, you're running it at
the point where the junction temperature of the transistor is kept to a
level that will let it stay a transistor for a satisfying amount of time.

This is why in my original reply I compared the output amp to a wall socket.
The equivalent Rs of your average wall socket is probably on the order 1/2
to 4 ohms (on the low end for 120V, high for 230). Since I live in the US
and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs
with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit
is only is only good for 15 amps and I'm asking for 120?!? Dang! What
happened to the lights?

So you see, every time you plug anything into the wall, you're failing to
match Rs with Rl, and that's a _very_ good thing. It's for exactly the same
reason that you don't match Rs with Rl in your output stage.

"Paul Burridge" wrote in message
...
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:
-- snip --
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
-
Winston Churchill

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Hi, Leon,

Do you know Brian (G4BCO) by any chance?
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

Paul Burridge wrote:
On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:


Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.


??? Shirley, some mistake.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html